How do you solve Single Element in a Sorted Array on LeetCode?

Single Element in a Sorted Array (LeetCode #540) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log n) time complexity and O(1) space complexity. Key insight: The array is sorted, which allows for efficient searching methods like binary search.

What is the brute force solution for Single Element in a Sorted Array?

The brute force approach for Single Element in a Sorted Array is Brute Force, with O(n) time complexity and O(1) space complexity. The brute force approach involves checking each element in the array and counting its occurrences. Since the array is sorted, we can simply iterate through the array and find the element that appears only once. The optimal Optimal Solution approach improves this to O(log n).

What algorithm or data structure is used to solve Single Element in a Sorted Array?

Single Element in a Sorted Array uses the following concepts: Array, Binary Search. The recommended patterns to study are: Binary Search, Array.

What are the interview tips for Single Element in a Sorted Array?

Always mention the time and space complexities of your approach. Consider edge cases and how your algorithm handles them. Practice explaining your thought process clearly as you code.

What are common mistakes when solving Single Element in a Sorted Array?

Not recognizing that the array is sorted and trying to use linear search instead of binary search. Failing to handle the edge case where the unique element is at the beginning or end of the array.

#540

Single Element in a Sorted Array

Medium

ArrayBinary SearchBinary SearchArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n)	O(log n)
Space	O(1)	O(1)

💡

Intuition

Time O(log n)Space O(1)

Using binary search takes advantage of the sorted nature of the array. By leveraging the properties of indices, we can efficiently narrow down the search space to find the unique element in O(log n) time.

⚙️

Algorithm

5 steps

1Step 1: Initialize left pointer to 0 and right pointer to the last index of the array.
2Step 2: While left is less than right, calculate the mid index.
3Step 3: Check if mid is even or odd and compare the mid element with its neighbor.
4Step 4: If they are the same, move the left pointer to mid + 1; otherwise, move the right pointer to mid.
5Step 5: When left equals right, return the element at that index.

solution.py14 lines

1# Full working Python code
2
3def single_non_duplicate(nums):
4    left, right = 0, len(nums) - 1
5    while left < right:
6        mid = left + (right - left) // 2
7        if mid % 2 == 1:
8            mid -= 1
9        if nums[mid] == nums[mid + 1]:
10            left = mid + 2
11        else:
12            right = mid
13    return nums[left]
14

ℹ

Complexity note: The time complexity is O(log n) because we are halving the search space with each iteration. The space complexity is O(1) as we are using a constant amount of space.

1The array is sorted, which allows for efficient searching methods like binary search.
2The unique element's position can be inferred from the properties of pairs.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.